A series for Euler's constant $\gamma$
Messenger of Mathematics, XLVI, 1917, 73 – 80
1. In a paper recently published in this Journal (Vol.XLIV, pp. 1 – 10), Dr. Glaisher proves a number of formulæ of the type $$\gamma = 1 - 2 \left(\frac{S_3}{3\cdot 4} + \frac{S_5}{5\cdot 6} + \frac{S_7}{7\cdot 8} + \cdots \right),$$ where $$ S_n = 1^{-n} + 2^{-n} + 3^{-n} + 4^{-n} + \cdots, $$ and conjectures the existence of a general formula $$ \gamma = \lambda_r - (r+1) (r+2) \cdots (2r)$$ $$\times \left\{\frac{S_3}{3(r+3)(r+4) \cdots(2r+2)} + \frac{S_5}{5(r+5)(r+6)\cdots(2r+4)} + \cdots \right\},$$ where $\lambda_r$ is a rational number. I propose now to prove the general formula of which Dr. Glaisher's are particular cases: this formula is itself a particular case of still more general formulæ.
2. Let $r$ and $t$ be any two positive numbers. Then
\begin{eqnarray}
&& \int\limits^1_0 x^{r-1} (1-x)^{t-1} \log \Gamma(1-x) \ dx = \int\limits^1_0
x^{t-1}(1-x)^{r-1} \log \Gamma (x) \ dx \nonumber \\
&& =\int\limits^1_0 x^{t-1} (1-x)^{r-1} \log \Gamma(1+x) \ dx
- \int\limits^1_0 x^{t-1}
(1-x)^{r-1} \log x \ dx
\end{eqnarray}
But
\begin{eqnarray}
&& \int\limits^1_0 x^{r-1} (1-x)^{t-1} \log \Gamma (1-x) \ dx \nonumber \\
&&= \int\limits^1_0 x^{r-1} (1-x)^{t-1} \left\{\gamma x + S_2\frac{x^2}{2}+S_3
\frac{x^3}{3} + \cdots \right\} \ dx \nonumber \\
&&=\frac{\Gamma(1+r)\Gamma(t)}{\Gamma(1+r+t)} \gamma + \frac{\Gamma(2+r)
\Gamma(t)}{\Gamma(2+r+t)}
\frac{S_2}{2}+\frac{\Gamma(3+r)\Gamma(t)}{\Gamma(3+r+t)}\frac{S_3}{3}+\cdots
\end{eqnarray}
Similarly
\begin{eqnarray*}
&& \int\limits^1_0 x^{t-1} (1-x)^{r-1} \log \Gamma(1+x) \ dx \\
&& = -\frac{\Gamma(1+t)\Gamma(r)}{\Gamma(1+r+t)} \gamma + \frac{\Gamma(2+t)
\Gamma(r)}{\Gamma(2+r+t)}
\frac{S_2}{2} -\frac{\Gamma(3+t)\Gamma(r)}{\Gamma(3+r+t)} \frac{S_3}{3}
+ \cdots \tag{2'}
\end{eqnarray*}
And also
\begin{eqnarray}
\int\limits^1_0 x^{t-1} (1-x)^{r-1} \log x \ dx &=&
\frac{d}{dt} \int\limits^1_0 x^{t-1}
(1-x)^{r-1} \ dx = \frac{d}{dt}
\left\{\frac{\Gamma(t)\Gamma(r)}{\Gamma(r+t)}\right\}\nonumber \\
&=&
\frac{\Gamma(r)\Gamma(t)}{\Gamma(r+t)}
\left\{\frac{\Gamma'(t)}{\Gamma(t)}-\frac{\Gamma'(r+t)}{\Gamma(r+t)}\right\}
\nonumber \\
&=& -\frac{\Gamma(r)\Gamma(t)}{\Gamma(r+t)} \int\limits^1_0 x^{t-1}
\frac{1-x^r}{1-x} \ dx
\end{eqnarray}
It follows from (1) – (3) that, if $r$ and $t$ are positive, then
\begin{eqnarray}
&& \frac{r}{1(r+t)}\gamma+ \frac{r(r+1)}{2(r+t)(r+t+1)} S_2+
\frac{r(r+1)(r+2)}{3(r+t)(r+t+1)(r+t+2)} S_3 + \cdots \nonumber \\
&&+ \frac{t}{1(r+t)}\gamma - \frac{t(t+1)}{2(r+t)(r+t+1)}S_2 +
\frac{t(t+1)(t+2)}{3(r+t)(r+t+1)(r+t+2)} S_3 - \cdots \nonumber \\
&&= \int\limits^1_0 \frac{x^{t-1}(1-x^r)}{1-x} \ dx
\end{eqnarray}
Now, interchanging $r$ and $t$ in (4), and taking the sum and the difference
of the two results, we see that, if $r$ and $t$ are positive, then
\begin{eqnarray}
&& \frac{r+t}{1(r+t)} \gamma +
\frac{r(r+1)(r+2)+t(t+1)(t+2)}{3(r+t)(r+t+1)(r+t+2)}S_3 + \cdots \nonumber \\
&&= \frac{1}{2} \int\limits^1_0 \frac{x^{r-1} + x^{t-1}
- 2x^{r+t-1}}{1-x} \ dx;
\end{eqnarray}
and
\begin{eqnarray}
\frac{r(r+1)-t(t+1)}{2(r+t)(r+t+1)}S_2
+\frac{r(r+1)(r+2)(r+3)-t(t+1)(t+2)(t+3)}{4(r+t)(r+t+1)(r+t+2)(r+t+3)} S_4 &&+ \cdots
\nonumber \\
&& = \frac{1}{2}\int\limits^1_0 \frac{x^{t-1} - x^{r-1}}{1-x} \ dx.
\end{eqnarray}
The right-hand sides of (5) and (6) can be expressed in finite terms if $r$
and $t$ are rational. If, in particular, $r$ and $t$ are integers, then
\begin{eqnarray*}
\int\limits^1_0 \frac{x^{r-1}+x^{t-1} - 2x^{r+t-1}}{1-x} \ dx &&= \frac{1}{r} +
\frac{1}{r+1} + \frac{1}{r+2} + \cdots +\frac{1}{r+t-1} \\
&&+ \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2} + \cdots + \frac{1}{r+t-1};
\end{eqnarray*}
and
\begin{eqnarray*}
\int\limits^1_0 \frac{x^{t-1}-x^{r-1}}{1-x} \ dx &&= \left(1+\frac{1}{2} + \frac{1}{3} +
\frac{1}{4} + \cdots + \frac{1}{r-1}\right)\\
&&- \left(1 + \frac{1}{2} + \frac{1}{3}+\frac{1}{4}+ \cdots +
\frac{1}{t-1}\right).
\end{eqnarray*}
3. Let us now suppose that $t=r$ in (5). Then it is clear that
\begin{eqnarray}
\gamma + \frac{(r+1)(r+2)}{3(2r+1)(2r+2)} S_3 &&+
\frac{(r+1)(r+2)(r+3)(r+4)}{5(2r+1)(2r+2)(2r+3)(2r+4)} S_5 + \cdots\nonumber \\
&&= \int\limits^1_0 \frac{x^{r-1}(1-x^r)}{1-x} \ dx = \int\limits^1_0
\frac{1+x^{2r-1}}{1+x} \ dx,
\end{eqnarray}
if $r > 0$. If we suppose, in (7), that $r$ is an integer, we obtain the
formula conjectured by Dr Glaisher, the value of $\lambda_r$ being
$$ \int\limits^1_0 \frac{1+x^{2r-1}}{1+x} \ dx = 1 - \frac{1}{2} + \frac{1}{3} -
\frac{1}{4} + \cdots + \frac{1}{2r-1}.$$
Again, dividing both sides in (6) by $r-t$ and making $t \to r$, we see that,
if $r > 0$, then
\begin{eqnarray}
&& \frac{r+1}{2(2r+1)} \left(\frac{1}{r} + \frac{1}{r+1}\right) S_2 \nonumber\\
&&+ \frac{(r+1)(r+2)(r+3)}{4(2r+1)(2r+2)(2r+3)} \left(\frac{1}{r} +
\frac{1}{r+1}+ \frac{1}{r+2} + \frac{1}{r+3}\right) S_4 + \cdots \nonumber \\
&&\qquad= - \int\limits^1_0 \frac{x^{r-1} \log x}{1-x} \ dx = \frac{1}{r^2}
+\frac{1}{(r+1)^2} + \frac{1}{(r+2)^2} + \frac{1}{(r+3)^2} +
\end{eqnarray}
Thus for example we have
$$ \frac{\pi^2}{12} = (1 + \myfrac{1}{2}) \frac{S_2}{2\cdots 3} + (1+\myfrac{1}{2}+
\myfrac{1}{3}+\myfrac{1}{4}) \frac{S_4}{4\cdot 5}
(1+\myfrac{1}{2} + \myfrac{1}{3}+
\myfrac{1}{4}+\myfrac{1}{5}+\myfrac{1}{6}) \frac{S_6}{6\cdot 7}+ \cdots.$$
4. If we start with the integral $$\int\limits^1_0 x^{r-1} (1-x)^{t-1} \log \Gamma \left(1 - \frac{x}{2}\right) \ dx,$$ and proceed as in $\S$ 2, we can shew that, if $r$ and $t$ are positive, then
\begin{eqnarray}
\frac{r}{1(r+t)}&& S'_1
+ \frac{r(r+1)}{2(r+t)(r+t+1)} S'_2
+ \frac{r(r+1)(r+2)}{3(r+t)(t+t+1)(r+t+2)} S'_3 + \cdots \nonumber \\
&&- \frac{t}{1(r+t)} S'_1 + \frac{t(t+1)}{2(r+t)(r+t+1)} S'_2
- \frac{t(t+1)(t+2)}{3(r+t)(r+t+1)(r+t+2)} S'_3 + \cdots \nonumber \\
&&= \int\limits^1_0 \frac{x^{t-1}(1-x^r)}{1-x} \ dx - \log \frac{\pi}{2},
\end{eqnarray}
where
$$S'_n = 1^{-n} - 2^{-n} + 3^{-n} - 4^{-n} + \cdots $$
From (9) we can easily deduce that, if $r$ and $t$ are positive, then
\begin{eqnarray}
&& \frac{r(r+1)+t(t+1)}{2(r+t)(r+t+1)} S'_2 \nonumber \\
&&+ \frac{r(r+1)(r+2)(r+3)+t(t+1)(t+2)(t+3)}{4(r+t)(r+t+1)(r+t+2)(r+t+3)}
S'_4 + \cdots\nonumber \\
&&= \frac{1}{2} \int\limits^1_0 \frac{x^{r-1} + x^{t-1} - 2x^{r+t-1}}{1-x} \ dx - \log
\frac{\pi}{2};
\end{eqnarray}
and
\begin{equation}
\frac{r-t}{1(r+t)} S'_1 +
\frac{r(r+1)(r+2)-t(t+1)(t+2)}{3(r+t)(r+t+1)(r+t+2)} S'_3 + \cdots
= \frac{1}{2} \int\limits^1_0 \frac{x^{t-1}- x^{r-1}}{1-x} \ dx .
\end{equation}
As particular cases of (10) and (11), we have
\begin{equation}
\log \frac{\pi}{2} + \frac{r+1}{2(2r+1)} S'_2 +
\frac{(r+1)(r+2)(r+3)}{4(2r+1)(2r+2)(2r+3)} S'_4 + \cdots
= \int\limits^1_0 \frac{1+x^{2r-1}}{1+x} \ dx;
\end{equation}
and
\begin{eqnarray}
\frac{1}{r} S'_1 &&+ \frac{(r+1)(r+2)}{3(2r+1)(2r+2)} \left(\frac{1}{r} +
\frac{1}{r+1} + \frac{1}{r+2} \right) S'_3 \nonumber \\
&&+ \frac{(r+1)(r+2)(r+3)(r+4)}{5(2r+1)(2r+2)(2r+3)(2r+4)} \nonumber \\
&&\times\left(\frac{1}{r}+ \frac{1}{r+1} + \frac{1}{r+2} + \frac{1}{r+3} +
\frac{1}{r+4} \right) S'_5 + \cdots \nonumber \\
&&= \frac{1}{r^2} + \frac{1}{(r+1)^2} + \frac{1}{(r+2)^2} + \cdots,
\end{eqnarray}
provided that $r \gt 0$. Thus for example we have
\begin{eqnarray*}
&& 1 = \log \frac{\pi}{2} + 2 \left(\frac{S'_2}{2.3} + \frac{S'_4}{4.5} +
\frac{S'_6}{6.7} + \cdots \right); \\
\frac{\pi^2}{12} = \frac{S'_1}{1.2} + \frac{S'_3}{3.4} &&(1+ \frac{1}{2} +
\frac{1}{3}) + \frac{S'_5}{5.6} (1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} +
\frac{1}{5}) + \cdots.
\end{eqnarray*}
5. The preceding results may be generalised as follows. Let $\zeta(s,x)$ denote the function represented by the series $$ x^{-s} + (x+1)^{-s} + (x+2)^{-s} + (x+3)^{-s} + \cdots (x \gt 0)$$ and its analytical continuations, so that $\zeta(s,1) = \zeta(s)$ and $\zeta(s, \frac{1}{2}) = (2^s - 1) \zeta(s), \zeta(s)$ being the Riemann $\zeta$-function. Then
\begin{eqnarray}
\int\limits^1_0 && x^{r-1} (1-x)^{t-1} \zeta(s,1-x) \ dx = \int\limits^1_0
x^{t-1} (1-x)^{r-1} \zeta(s,x) \ dx \nonumber \\
&&= \int\limits^1_0 x^{t-1} (1-x)^{r-1} \zeta(s,1 + x) \ dx + \int\limits^1_0
x^{t-s-1}(1-x)^{r-1} \ dx,
\end{eqnarray}
provided that $r$ and $t$ are positive. But we know that, if $|x|\lt 1$, then
\begin{equation}
\zeta(s,1 -x) = \zeta(s) + \frac{s}{1!} \zeta(s+1) x + \frac{s(s+1)}{2!}
\zeta(s+2) x^2 + \cdots;
\end{equation}
and that
\begin{equation}
\int\limits^1_0 x^{t-s-1} (1-x)^{r-1} \ dx =
\frac{\Gamma(t-s)\Gamma(r)}{\Gamma(r-s+t)},
\end{equation}
provided that $t > s$. It follows from (14)–(16) that, if $r$ and $t$ are
positive and $t > s$, then
\begin{eqnarray}
&& \left\{\zeta(s) + \frac{s}{1!} \frac{r}{r+t} \zeta(s+1) +
\frac{s(s+1)}{2!} \frac{r(r+1)}{(r+t)(r+t+1)} \zeta(s+2) + \cdots\right\}\nonumber \\
&&- \left\{\zeta(s) - \frac{s}{1!}\frac{t}{r+t} \zeta(s+1) + \frac{s(s+1)}{2!}
\frac{t(t+1)}{(r+t)(r+t+1)} \zeta(s+2) - \cdots \right\}\nonumber \\
&&= \frac{\Gamma(r+t) \Gamma(t-s)}{\Gamma(t) \Gamma(r-s+t)}.
\end{eqnarray}
As particular cases of (17), we have
\begin{eqnarray}
\frac{s}{1!}\frac{r+t}{r+t} \zeta(s+1) + \frac{s(s+1)(s+2)}{3!} &&
\frac{r(r+1)(r+2)+t(t+1)(t+2)}{(r+t)(r+t+1)(r+t+2)}
\zeta(s+3) + \cdots \nonumber \\
&&= \frac{1}{2} \frac{\Gamma(r+t)}{\Gamma(r-s+t)}
\left\{\frac{\Gamma(t-s)}{\Gamma(t)} + \frac{\Gamma(r-s)}{\Gamma(r)} \right\},
\end{eqnarray}
and
\begin{eqnarray}
\frac{s(s+1)}{2!} && \frac{r(r+1)-t(t+1)}{(r+t)(r+t+1)} \zeta(s+2)\nonumber \\
&&+\frac{s(s+1)(s+2)(s+3)}{4!}
\frac{r(r+1)(r+2)(r+3)-t(t+1)(t+2)(t+3)}{(r+t)(r+t+1)(r+t+2)(r+t+3)}
\zeta(s+4)+ \cdots \nonumber \\
&&= \frac{1}{2} \frac{\Gamma(r+t)}{\Gamma(r-s+t)}
\left\{\frac{\Gamma(t-s)}{\Gamma(t)} - \frac{\Gamma(r-s)}{\Gamma(r)}\right\},
\end{eqnarray}
provided that $r$ and $t$ are positive and greater than $s$. From (18) and
(19) we deduce that, if $r$ is positive and greater than $s$, then
\begin{eqnarray}
\frac{s}{1!} \frac{r}{2r} \zeta(s+1) + \frac{s(s+1)(s+2)}{3!}
\frac{r(r+1)(r+2)}{2r(2r+1)(2r+2)} && \zeta(s+3) + \cdots\nonumber \\
= \frac{1}{2} \frac{\Gamma(2r)\Gamma(r-s)}{\Gamma(r)\Gamma(2r-s)},
\end{eqnarray}
and
\begin{eqnarray}
&& \frac{s(s+1)}{2!} \frac{r(r+1)}{2r(2r+1)} \left(\frac{1}{r} +
\frac{1}{r+1} \right)\zeta(s+2)\nonumber \\
&& +\frac{s(s+1)(s+2)(s+3)}{4!} \frac{r(r+1)(r+2)(r+3)}{2r(2r+1)(2r+2)(2r+3)}\nonumber \\
&& \times \left(\frac{1}{r} + \frac{1}{r+1} + \frac{1}{r+2} +
\frac{1}{r+3}\right) \zeta(s+4) + \cdots \nonumber \\
&& = \frac{1}{2} \frac{\Gamma(2r)\Gamma(r-s)}{\Gamma(r)\Gamma(2r-s)}
\int\limits^1_0 \frac{x^{r-s-1}(1-x^s)}{1-x} \ dx.
\end{eqnarray}
6. If we start with the integral $$ \int\limits^1_0 x^{r-1} (1-x)^{t-1} \zeta \left(s,1 - \frac{x}{2}\right) \ dx,$$ and proceed as in $\S$ 5, we can shew that, if $r$ and $t$ are positive and $t > s$, then
\begin{eqnarray}
\zeta_1 (s) && + \frac{s}{1!} \frac{r}{r+t} \zeta_1(s+1) \frac{s(s+1)}{2!}
\frac{r(r+1)}{(r+t)(r+t+1)}\zeta_1(s+2)+ \cdots\nonumber \\
&& + \zeta_1 (s) - \frac{s}{1!} \frac{t}{r+t} \zeta_1(s+1) + \frac{s(s+1)}{2!}
\frac{t(t+1)}{(r+t)(r+t+1)} \zeta_1 (s+2) - \cdots \nonumber \\
&& = \frac{\Gamma(r+t)\Gamma(t-s)}{\Gamma(t)\Gamma(r-s+t)},
\end{eqnarray}
where $\zeta_1(s)$ is the function represented by the series
$$1^{-s} - 2^{-s} + 3^{-s} - 4^{-s} + \cdots $$
and its analytical continuations. From (22) we deduce that, if $r$ and $t$
are positive and greater than $s$, then
\begin{eqnarray}
(1+1) \zeta_1(s) &&+ \frac{s(s+1)}{2!} \frac{r(r+1)+t(t+1)}{(r+t)(r+t+1)}
\zeta_1(s+2) + \cdots \nonumber \\
&&= \frac{1}{2} \frac{\Gamma(r+t)}{\Gamma(r-s+t)}
\left\{\frac{\Gamma(t-s)}{\Gamma(t)} + \frac{\Gamma(r-s)}{\Gamma(r)}\right\};
\end{eqnarray}
and
\begin{eqnarray}
&&\frac{s}{1!} \frac{r-t}{r+t} \zeta_1 (s+1)\nonumber \\
&+& \frac{s(s+1)(s+2)}{3!}
\frac{r(r+1)(r+2)-t(t+1)(t+2)}{(r+t)(r+t+1)(r+t+2)}
\zeta_1(s+3) + \cdots \nonumber \\
&=& \frac{1}{2} \frac{\Gamma(r+t)}{\Gamma(r-s+t)}
\left\{\frac{\Gamma(t-s)}{\Gamma(t)} - \frac{\Gamma(r-s)}{\Gamma(r)}\right\}.
\end{eqnarray}
As particular cases of (23) and (24), we have
\begin{equation}
\zeta_1(s) + \frac{s(s+1)}{2!} \frac{r(r+1)}{2r(2r+1)} \zeta_1(s+2) +
\cdots = \frac{1}{2} \frac{\Gamma(2r)\Gamma(r-s)}{\Gamma(r) \Gamma(2r-s)},
\end{equation}
and
\begin{eqnarray}
\frac{s}{1!} \frac{r}{2r} \frac{1}{r} \zeta_1(s+1)
&+& \frac{s(s+1)(s+2)}{3!} \frac{r(r+1)(r+2)}{2r(2r+1)(2r+2)}
\left(\frac{1}{r}
+ \frac{1}{r+1} + \frac{1}{r+2} \right) \zeta_1 (s+3) + \cdots\nonumber \\
&=&\frac{1}{2} \frac{\Gamma(2r)\Gamma(r-s)}{\Gamma(r)\Gamma(2r-s)}
\int\limits^1_0 \frac{s^{r-s-1}(1-x^s)}{1-x} \ dx,
\end{eqnarray}
provided that $r$ is positive and greater than $s$.